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2 years ago

As the mass of object increased it is density increased

Physics

1 answer:

Elan Coil [88]2 years ago

3 0

Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

$d=\dfrac{m}{V}$

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.

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2 years ago

A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

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3 years ago

Differentiate the following functions with respect to x <br>xsin x

gtnhenbr [62]

Answer:

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3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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2 years ago

You are standing on a skateboard, initially at rest. a friend throws a very heavy ball towards you. you can either catch the obj

PIT_PIT [208]

<span>You should deflect the ball in order to maximize your speed on the skateboard.

Since this creates a larger impulse, you want to deflect the ball. Splitting it up into catching and throwing the ball may by something you can think of deflecting the ball. First, you need to catch the ball, which in turn would push you forward with some speed. (The speed we are talking about should obviously be equal to option A, where you catch the ball). Now, throw the ball back to him since these two processes are equal to deflecting the ball. Throwing a mass away from you would cause or enable you to move even fast.</span>

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2 years ago